A131401 Least number dividing Fibonacci(n) but not dividing Fibonacci(m) for m < n, or 0 if there is no such number.

1, 0, 2, 3, 5, 4, 13, 7, 17, 11, 89, 6, 233, 29, 10, 47, 1597, 19, 37, 15, 26, 199, 28657, 14, 25, 521, 53, 39, 514229, 20, 557, 2207, 178, 3571, 65, 27, 73, 9349, 466, 35, 2789, 52, 433494437, 43, 85, 139, 2971215073, 64, 97, 101, 3194, 699, 953, 212, 445, 49, 74, 59

Offset: 1

Author

Herbert A. Hauptman (hauptman(AT)hwi.buffalo.edu) & Robert G. Wilson v, Jul 07 2007

Keywords

• nonn

Comments

First occurrence of n in A001177 or 0 if impossible.

Conjecture: only a(2)=0. I have not found values of a(n) < 2*106 less than 100 for n = 43, 47, 74, 82, 83 & 94.

When Fibonacci(n) is a prime number, then a(n)=Fibonacci(n). Note that a(n)=0 for n=2 because Fibonacci(1)=Fibonacci(2)=1. For n > 2, an upper bound for a(n) is Fibonacci(n). The difficulty in computing this sequence for large n is the factorization of Fibonacci(n), which is required to find the divisors of Fibonacci(n). - T. D. Noe, Jan 12 2009

In other words, the conjecture is true. For n > 2, Fibonacci(n) has at least one divisor that does not divide Fibonacci(k) for k < n. The number of such divisors is A120256(n).

References

• Alfred S. Posamentier & Ingmar Lehmann, The (Fabulous) Fibonacci Numbers, Afterword by Herbert A. Hauptman, 2. 'The Minor Modulus m(n)', Prometheus Books, NY, 2007, pp. 329-342.

Links

• T. D. Noe, Table of n, a(n) for n = 1..300

Crossrefs

Cf. A060442. - T. D. Noe, Jan 12 2009

Programs

Mathematica

f[n_] := Block[{k = 1}, While[Mod[Fibonacci@k, n] != 0 && k < 101, k++ ]; k]; t = Table[0, {100}]; Do[ a = f@n; If[a < 101 && t[[a]] == 0, t[[a]] = n; Print[{a, n}]], {n, 106}]
nn=100; fib=Fibonacci[Range[nn]]; Join[{1,0}, Table[dvrs=Rest[Divisors[fib[[n]]]]; k=1; While[d=dvrs[[k]]; pos=Position[fib,?(Mod[ #,d]==0&),1,1]; pos!={{n}}, k++ ]; d, {n,3,nn}]] (* _T. D. Noe, Jan 12 2009 *)

Extensions

Extended by T. D. Noe, Jan 12 2009